It's easy if you use Z[sqrt(-2)] and Z[sqrt(2)] (they're both UFD). For example look at a^2 + 2 = b^3. This factors as (a + sqrt(-2))(a - sqrt(-2)) = b^3, and the units of Z[sqrt(-2)] are just +/-1, so a + sqrt(-2) = (c + d*sqrt(-2))^3 for some c,d in Z. So a = c^3 - 6cd^2 = c*(c^2 - 6d^2) and 1 = d*(3c^2 - 2d^2). So d = +/-1 and c = +/- 1, whence you get a = +/-5, and so b = 2. Same situation for a^2 - 2 = b^3.
Seem O.K. to me, except b=3. In the second case we have a=+/-1 b=-1. However, Dr. Math makes it more complicated: http://mathforum.org/library/drmath/view/51569.html
Yea, I don't know why the post in that link is using both a + sqrt(-2) = w^3 and a - sqrt(-2) = t^3 (conjugate). You only need the first, you get just as much information just more quickly.
To go from [tex](a + \sqrt{-2})(a - \sqrt{-2}) = b^3[/tex] to [tex]a + \sqrt{-2} = (c + d\sqrt{-2})^3[/tex] don't we need that [itex]a + \sqrt{-2}[/itex] and [itex]a - \sqrt{-2}[/itex] are relatively prime? In fact, as Doctor Rob points out on the Math Forum, GCD([itex]a + \sqrt{-2}, a - \sqrt{-2}[/itex]) divides [itex]2\sqrt{-2}[/itex]. I think that's why he has to consider several cases.
Yes, that's right. So what I wrote above is not entirely correct, but one gets the idea - just factor and use UFD-ness of Z[sqrt(-2)]. More importantly, does anyone know how to use tex here without having to type itex and \itex everytime I want it? for example i prefer to $ $, just much easier.
The pdf file linked at the beginning of this thread strongly suggests that itex, /itex and its variants are the only format that works here.
Hi folks, I followed a different direction. first notice that a and b must have the same parity, so first case: 1) a, b = even [tex](2k)^{2}\pm 2 = (2n)^{3}[/tex] [tex]4k^{2}\pm 2 = 8n^{3}[/tex] so, [tex]2k^{2}\pm 1= 4n^{3}[/tex] absurd, since even number cannot be equal to odd number, so second case: 2) a, b = odd [tex](2k+1)^{2}\pm 2 = (2n+1)^{3}[/tex] 2.1) + 2 [tex]4k(k+1)+3 = 8n^{3}+12n^{2}+6n+1[/tex] wich means [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex] [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex] and [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (K+1)[/tex] so [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex] [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex] and [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (K+1)[/tex] given [tex]f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + .... + a_{i}x^{m_{i}}[/tex] and [tex]g(x) = b_{0} + b_{1}x + b_{2}x^{2} + b_{3}x^{3} + ....+ b_{j}x^{n_{j}}[/tex] we know that [tex]f(x)\equiv\ g(x)\mod W[/tex] [tex]\Leftrightarrow[/tex] [tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex] lets make [tex]f(x) = 4n^{3}+6n^{2}+3n[/tex] and [tex]g(x) = 1[/tex] finally (skip obvious manipulations) [tex]12\equiv\ 0 \mod 4[/tex] [tex]12\equiv\ 0 \mod k[/tex] and [tex]12\equiv\ 0 \mod (K+1)[/tex] k must be = 2 2.2) - 2 (...) (after the same steps) finally [tex]14\equiv\ 0 \mod 4[/tex] what is untrue Any remarks would be apreciated.
Thinking again, I suspect I misunderstood the theorem found here: http://planetmath.org/encyclopedia/PolynomialCongruence.html and jumped to a conclusion prematurely. [tex]f(x)\equiv\ g(x)\mod W[/tex] [tex] a_{0}+a_{1}+a_{2}+a_{3}+ .... +a_{i}\equiv\ b_{0}+b_{1}+b_{2}+b_{3}+ ....+b_{j}\ mod W[/tex] Seems that the conclusion above is wrong. Anyway, I'd preciate if anyone could make the theorem more clear to me, and make any comments about what I've posted :) thank you!
[tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod 4[/tex] [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod k[/tex] and [tex]8n^{3}+12n^{2}+6n+1\equiv\ 3 \mod (k+1)[/tex] so [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod 4[/tex] [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod k[/tex] and [tex]4n^{3}+6n^{2}+3n\equiv\ 1 \mod (k+1)[/tex] that's also wrong, but do not detroy the following argument because [tex]24\equiv\ 0 \mod 4[/tex] [tex]24\equiv\ 0 \mod k[/tex] and [tex]24\equiv\ 0 \mod (k+1)[/tex] still would imply that k must be = 2